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Posted

how about 15 +15 ??

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Posted

But you need 3 numbers...

Isaiah 32:17 And the work of righteousness shall be peace; and the effect of righteousness quietness and assurance for ever.

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Posted

how about 15 +15 ??

I was thinking the same, but than read where it said, fill the boxes!! On second thought, it doesn't say that you have to fill "all" the boxes??

phkrause

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Posted

Touché!

Isaiah 32:17 And the work of righteousness shall be peace; and the effect of righteousness quietness and assurance for ever.

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Posted

I think it's mathematically insoluble, as set up. Adding three odd numbers together will never yield an even number.

If it was possible to add a minus sign, it would be easy...

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Truth is important

Posted

I think it's mathematically insoluble, as set up. Adding three odd numbers together will never yield an even number.

If it was possible to add a minus sign, it would be easy...

 

 

I would assume that the solution was that it was unsolvable. 

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Posted

It cannot be solved using whole numbers, but it can with rational numbers, for example:

 

1.5 + 15.5 + 13

 

There are various other combinations that will also work.

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Posted

Box 1 -   15

Box 2 - 13

Box 3 -  1,1,

 

?

If that is OK to do??

phkrause

When the righteous are in authority, the people rejoice; But when a wicked man rules, the people groan. Proverbs 29;2
  • Moderators
Posted

Or:  11,13, 7-1= 30

Gregory

Posted

9,9 + 7 + 5 = 30

 

Is only one solution, it says you can fill the boxes and that you can repeat the numbers. So there are multiple solutions to this problem. :)

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Posted

Those solutions all imply an additional + or - sign within a box that isn't in evidence, so I think they're pretty shaky, even if we allowed more than one number per box.

Truth is important

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Posted

Someone got me thinking - we're assuming these are normal base-10 numbers. Perhaps in some other number base it can be solved?

If it is in 'base-x', then '30' is just 3x. Single digit numbers are just themselves, and double digit numbers are x+{the second digit} since they only go to 1 in the left-most digit.

So, for example, if it's in base-3, the answer on the right is 3x3 = 9 in decimal.

1 + 3 + '11' (which is 4 in decimal) = 8 in decimal, so it misses, and if we go to '13' (6 in decimal) it goes to 10 as the answer and still misses.

Given that the number '15' can occur (as stated in the question), it's likely to be at least base-6 (digits 0 to 5), and maybe an even number will work better than an odd.

In base-6, '30' is 18 in decimal.

3 + 7 + '11' (which is 7 in decimal) = 17 in decimal

Hmm - still getting odd numbered answers from the left hand side, needing even numbered answers on the right hand side and still being able to only increment and decrement the left hand side in twos.

Base-7 will give us an odd-numbered right hand side: '30' is 21 in decimal.

5 + 7 + '11' (which is 8 in decimal) = 20... d'oh! Now we have odd on the right and even on the left, and can *still* only jump in twos!

We'll, I'm defeated there, but maybe my struggles will give someone else a helpful idea.

Truth is important

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Posted (edited)

I think it's mathematically insoluble, as set up. Adding three odd numbers together will never yield an even number.

If it was possible to add a minus sign, it would be easy...

​Oh, give it a few billion years and that might do it.  (Sorry, Bravus, I couldn't resist that one. :))

Edited by Gerry Cabalo
mis-spelled a word.

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